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This question is still waiting for a better alternate answer. If you can add it then your answer can be chosen as accepted


I came across a question yesterday

A container contains three gases. A, B and C in equilibrium $\ce{A <=> 2B + C}$.
At equilibrium the concentration of A was 3M, and of B was 4M. On doubling the volume of the container, the new equilibrium concentration of B was 3M. Calculate $K_c$ and initial equilibrium concentration of C.


My approach:

                              A      <=>      2B     +     C
initial molarity             3M               4M           x
initial number of moles      3V               4V           xV
final molarity                y               3M           z
final number of moles        y*(2V)           6V           z*(2V)

Here $V$ is volume.

Now as there is $2V$ increase in the number of moles of $B$, so there should be V decrease in the number of moles of $A$. So $$y(2V) = 3V-V=2V$$ $$\Rightarrow y= 1M$$

Similarly there is V increase in the number of moles of $C$ $$z(2V)=(x+1)V$$$$\Rightarrow z=\dfrac{x+1}{2}$$

As equilibrium constant is same in both situations. So, $$\dfrac{4^2x}{3}=\dfrac{3^2\dfrac{x+1}{2}}{1}$$$$\Rightarrow x=27/5M$$

After finding $x$ we can easily find $K_c$ which is found to be $28.8 M^2$


My Question: Is there any easy approach that can help me to do these kind of questions involving stoichiometric coefficients in a bit quicker way? I am having a hard time with these stoichiometric coefficients. So the only thing that was hard to find (for me) was concentration of C. After finding its concentration $K_c$ was a piece of cake. It took me a lot of time to get to this correct solution.

Thanks in advance

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I do not think that there's a much simpler approach to solve the problem.

  • The first condition is that the number of moles $a + b/2$ is constant.
  • The second condition is that the number of moles $a + c$ is constant.
  • The third condition is that $k_c = \frac{1}{V^2} \frac{b^2 c}{a}$.

From the first condition you can derive the number of moles of A in the new equilibrium. From the second condition you can derive the difference in the number of moles of C between the two equilibria. From the third condition you can get the quotient thereof, and from both you can calculate the number of moles of C in the first equilibrium. Then you can calculate $K_c$.

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  • $\begingroup$ What are a, b and c here...? $\endgroup$ – manshu Feb 1 '16 at 20:51
  • $\begingroup$ The number of moles of A, B, and C respectively. $\endgroup$ – aventurin Feb 1 '16 at 21:08
  • $\begingroup$ +1 but you just wrote my own answer in short. I was looking for an alternate answer or some short trick. $\endgroup$ – manshu Feb 1 '16 at 21:15
  • $\begingroup$ This is why I wrote "I do not think that there's a much simpler approach to solve the problem". The math reflects the complexity of the problem. So it cannot be simplified arbitrarily. ;-) $\endgroup$ – aventurin Feb 1 '16 at 21:36

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